The straight line along with the point are important elements of geometry, with the help of which many figures are constructed in space and on the plane. This article describes in detail the parametric equation of the line, as well as its relationship with other types of equations for this geometric element.
Line and equations for its description
A straight line in geometry is a collection of points that connect two arbitrary points of space with a segment with the shortest length. This segment is part of the line. Any other curves connecting two fixed points in space will have a large length, therefore they are not straight.
The figure above shows two black dots. The blue line connecting them is a straight line, and the red line is a curve. Obviously, the length of the red line between the black dots is longer than the blue.
There are several types of equations of the line, with which you can describe the line in three-dimensional space or in two-dimensional. The following are the names of these equations:
- vectorial;
- parametric;
- in segments;
- symmetric or canonical;
- general type.
In this article, we consider the parametric equation of the line, however, we derive it from the vector. We also show the connection of parametric and symmetric or canonical equations.
Vector equation
It is clear that all the above types of equations for the considered geometric element are interconnected. Nevertheless, the vector equation is basic for all of them, since it directly follows from the definition of a line. Consider how it is introduced into geometry.
Suppose we are given a point in the space P (x 0 ; y 0 ; z 0 ). It is known that this point belongs to a line. How many lines can you draw through it? Endless variety. Therefore, in order to be able to draw a single line, you must specify the direction of the latter. The direction, as you know, is determined by the vector. We denote it by v¯ (a; b; c), where the characters in brackets are its coordinates. For each point Q (x; y; z), which is located on the line under consideration, we can write the equality:
(x; y; z) = (x 0 ; y 0 ; z 0 ) + α × (a; b; c)
Here, the symbol α is a parameter that takes absolutely any real value (multiplying a vector by a number can only change its modulus or direction in the opposite). This equality is called the vector equation for a line in three-dimensional space. Changing the parameter α, we get all the points (x; y; z) that form this line.
The vector v¯ (a; b; c) in the equation is called the directing vector. A straight line has no specific direction, and its length is infinite. These facts mean that any vector obtained from v¯ by multiplying by a real number will also be a guide for the line.
As for the point P (x 0 ; y 0 ; z 0 ), instead of it, we can substitute an arbitrary point in the equation that lies on a straight line, and the latter will not change.
The figure above shows the straight line (blue line), which is defined in space through the guide vector (red directional segment).
It is not difficult to obtain such equality for the two-dimensional case. Using similar reasoning, we arrive at the expression:
(x; y) = (x 0 ; y 0 ) + α × (a; b)
We see that it is completely the same as the previous one, only two coordinates are used instead of three to specify points and vectors.
The equation is parametric
First, we obtain in space the parametric equation of the line. Above, when vector equality was written, the parameter that is present in it was already mentioned. To get a parametric equation, just open the vector. We get:
x = x 0 + α × a;
y = y 0 + α × b;
z = z 0 + α × c
The combination of these three linear equalities, each of which has one variable coordinate and parameter α, is called the parametric equation of a straight line in space. In fact, we did not do anything new, but simply explicitly wrote down the meaning of the corresponding vector expression. We note only one point: although the number α is arbitrary, it is the same for all three equalities. For example, if α = -1.5 for the 1st equality, then its same value should be substituted into the second and third equalities when determining the coordinates of a point.
The parametric equation of a straight line on a plane is similar to that for the spatial case. It is written as:
x = x 0 + α × a;
y = y 0 + α × b
Thus, in order to compose a parametric equation of a line, we must explicitly write a vector equation for it.
Obtaining the canonical equation
As noted above, all equations defining a straight line in space and on the plane are obtained from one another. We show how to obtain a canonical line from a parametric equation. For the spatial case we have:
x = x 0 + α × a;
y = y 0 + α × b;
z = z 0 + α × c
Express the parameter in each equality:
α = (x - x 0 ) / a;
α = (y - y 0 ) / b;
α = (z - z 0 ) / c
Since the left sides are the same, then the right sides of the equalities are also equal to each other:
(x - x 0 ) / a = (y - y 0 ) / b = (z - z 0 ) / c
This is the canonical equation for a straight line in space. The denominator value in each expression is the corresponding coordinate of the guide vector. The values in the numerator that are subtracted from each variable represent the coordinates of a point that belongs to this line.
The corresponding equation for the case on the plane will take the form:
(x - x 0 ) / a = (y - y 0 ) / b
Further in the article we will solve several problems using the knowledge gained.
Equation of a line through 2 points
It is known that two fixed points both on a plane and in space uniquely define a straight line. Suppose you are given the following two points on the plane:
P (x 1 ; y 1 );
Q (x 2 ; y 2 )
How to make a straight line equation through them? First you need to determine the direction vector. Its coordinates have the following meanings:
PQ¯ (x 2 - x 1 ; y 2 - y 1 )
Now you can write the equation in any of the three types that were discussed in the paragraphs above. For example, the parametric equation of the line takes the form:
x = x 1 + α × (x 2 - x 1 );
y = y 1 + α × (y 2 - y 1 )
In canonical form, you can rewrite it like this:
(x - x 1 ) / (x 2 - x 1 ) = (y - y 1 ) / (y 2 - y 1 )
It can be seen that the coordinates of both points are included in the canonical equation, and these points can be changed in the numerator. So, the last equation can be rewritten as follows:
(x - x 2 ) / (x 2 - x 1 ) = (y - y 2 ) / (y 2 - y 1 )
All written expressions are called equations of the line through 2 points.
Three-point problem
The coordinates of the following three points are given:
M (5; 3; -1);
N (2; 2; 0);
K (1; -1; -5)
It is necessary to determine whether these points lie on one line or not.
This problem should be solved as follows: first, draw up the equation of the line for any two points, and then substitute the coordinates of the third one into it and check whether they satisfy the obtained equality.
We make the equation through M and N in parametric form. To do this, we apply the formula obtained in paragraph above, which we generalize to the three-dimensional case. We have:
x = 5 + α × (-3);
y = 3 + α × (−1);
z = -1 + α × 1
Now we substitute the coordinates of the point K in these expressions and find the value of the alpha parameter that corresponds to them. We get:
1 = 5 + α × (-3) => α = 4/3;
-1 = 3 + α × (-1) => α = 4;
-5 = -1 + α × 1 => α = -4
We found that all three equalities are true if each of them takes a different value of the parameter α. The latter fact contradicts the condition of the parametric equation of the line in which α must be equal for all equations. This means that the point K of the line MN does not belong, which means that all three points on the same line do not lie.
The problem of parallel lines
Two equations of lines are given in parametric form. They are presented below:
x = -1 + 5 × α;
y = 3 + 3 × α
and
x = 2 - 6 × λ;
y = 4 - 3.6 × λ
It is necessary to determine whether the lines are parallel. The easiest way is to determine the parallelism of two lines using the coordinates of the direction vectors. Turning to the general formula of the parametric equation in two-dimensional space, we find that the direction vectors of each line will have the coordinates:
v 1 ¯ (5; 3);
v 2 ¯ (-6; -3.6)
Two vectors are parallel if one of them can be obtained by multiplying the other by a certain number. We divide the coordinates of the vectors in pairs, we get:
-6/5 = -1.2;
-3.6 / 3 = -1.2
It means that:
v 2 ¯ = -1.2 × v 1 ¯
The directing vectors v 2 ¯ and v 1 ¯ are parallel, which means that the lines in the condition of the problem are also parallel.
Check if they are the same line. To do this, substitute the coordinates of any point in the equation for another. Take the point (-1; 3), substitute it in the equation for the second line:
-1 = 2 - 6 × λ => λ = 1/2;
3 = 4 - 3.6 × λ => λ ≈ 0.28
That is, the lines are different.
The perpendicularity problem of straight lines
Equations of two lines are given:
x = 2 × α;
y = 1 + 3 × α
and
x = 2 + 6 × λ;
y = -2 - 4 × λ
Are these lines perpendicular?
Two lines will be perpendicular if the scalar product of their guide vectors is zero. We write these vectors:
v 1 ¯ (2; 3);
v 2 ¯ (6; -4)
Find their scalar product:
(v 1 ¯ × v 2 ¯) = 2 × 6 + 3 × (-4) = 12 - 12 = 0
Thus, we found that the lines considered are perpendicular. They are shown in the figure above.